A barrel has 100 litres of pure alcohol. 20 litres of alcohol is removed and replaced with the same amount of water. This process is repeated two more times. What is the amount of alcohol in the resultant solution?
- a)40 litres
- b)36 litres
- c)51.2 litres
- d)54.6 litres
Correct answer is option 'C'. Can you explain this answer?
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Answers
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Ayashkant Dwibedy
Mar 29, 2020 |
Hey aarushi, there's a simple formula to this. Let a be the quantity of alcohol, b be the quantity removed and n be the no. of times operation is performed. Then,
fraction of original liquid(alcohol) left=(a-b/a) raised to the power n times.
Here, fraction of alcohol left= (100-20/100) to the power 3= 0.512
0.512 of 1= 51.2% of original amount.
Hence answer is c)
fraction of original liquid(alcohol) left=(a-b/a) raised to the power n times.
Here, fraction of alcohol left= (100-20/100) to the power 3= 0.512
0.512 of 1= 51.2% of original amount.
Hence answer is c)
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Wizius Careers
Feb 18, 2022 |
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The above solution of alcohol has been successively diluted three times. Hence we use the formula for successive dilution.
Where v is the initial volume of pure alcohol, c is the concentration of the alcohol after replacements and x is the volume of alcohol replaced.
51.2 litres
Hence, option 3.
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Spiritual Rasoi
Jan 15, 2022 |
We can use the formula
Quantity left = Quantity present initially * ( 1 - quantity removed/ original quantity) raise to power n.
where n = number of iterations.
Quantity left = 100 * ( 1 - 20/100) ^ 3 = 51.2litres
Quantity left = Quantity present initially * ( 1 - quantity removed/ original quantity) raise to power n.
where n = number of iterations.
Quantity left = 100 * ( 1 - 20/100) ^ 3 = 51.2litres
Hey aarushi, there's a simple formula to this. Let a be the quantity of alcohol, b be the quantity removed and n be the no. of times operation is performed. Then,fraction of original liquid(alcohol) left=(a-b/a) raised to the power n times. Here, fraction of alcohol left= (100-20/100) to the power 3= 0.5120.512 of 1= 51.2% of original amount.Hence answer is c)
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Hey aarushi, there's a simple formula to this. Let a be the quantity of alcohol, b be the quantity removed and n be the no. of times operation is performed. Then,fraction of original liquid(alcohol) left=(a-b/a) raised to the power n times. Here, fraction of alcohol left= (100-20/100) to the power 3= 0.5120.512 of 1= 51.2% of original amount.Hence answer is c)
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